Question

5.
A ball is thrown vertically upward with initial velocity 30 m/sec. What will be its position vector at
time t = 5 sec, taking origin at 45 m above the point of projection, vertical up as positive y-axis and
horizontal as x-axis :-
Min 051
(C) ( 15)
(
D(O ,5)
(
D(O ,-20)
​

Answer 1

answer : option (D) (0, -20)

explanation : A ball is thrown vertically upward with initial velocity 30m/sec. at maximum height , velocity of ball becomes zero.

so, using formula, v = u + at

0 = 30 - 10t

or, t = 3 sec

hence, ball becomes rest after 3 sec.

let's find height covered by ball , h

h = ut + 1/2 at²

= 30 × 3 - 1/2 × 10 (3)²

= 90 - 45

= 45m.

hence, body is at 45m above the projection after 3 sec. now ball starts to fall and after 2sec , ball falls , s = 1/2 × 10(2)² = 20m

so, final position of ball after 5 sec = 45 - 20 = 25m above the projection point.

it is given that, origin is at 45m above the projection point.

so, location of ball in vertical direction, y = 25 - 45 = -20m

and location of ball in horizontal direction, x = 0

so, position of ball = (0, -20)